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UPSC CDS Math Basic operation and Factorisation Questions

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Question : 77 of 77

Marks: +1, -0

If

\left(x^{2}+\frac{1}{x^{2}}\right)=\frac{17}{4}

\left(x^{3}-\frac{1}{x^{3}}\right)

$\;\frac{75}{16}$
$\;\frac{63}{8}$
$\;\frac{95}{8}$
None of these

Solution:

Given,

\left(x^{2}+\frac{1}{x^{2}}\right)=\frac{17}{4}

\Rightarrow x^{2}+\frac{1}{x^{2}}+2-2=\frac{17}{4}

\Rightarrow \;\; \left(x-\frac{1}{x}\right)^{2}+2=\frac{17}{4}

\Rightarrow \;\; \left(x-\frac{1}{x}\right)^{2}=\frac{17}{4}-2

\Rightarrow \;\; \left(x-\frac{1}{x}\right)^{2}=\frac{9}{4}

\Rightarrow \;\; \left(x-\frac{1}{x}\right)=\frac{3}{2}

[

\because

taking square root of both sides]

On cubing both sides, we get

\& \left(x-\frac{1}{x}\right)^{3}=\left(\frac{3}{2}\right)^{3}

\Rightarrow x^{3}-\frac{1}{x^{3}}-3 \times \frac{1}{x} \cdot x\left(x-\frac{1}{x}\right)=\frac{27}{8}

\Rightarrow x^{3}-\frac{1}{x^{3}}=\frac{27}{8}+3 \times \frac{3}{2}

\Rightarrow x^{3}-\frac{1}{x^{3}}=\frac{27}{8}+\frac{9}{2}

\therefore \; x^{3}-\frac{1}{x^{3}}=\frac{63}{8}

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